Two charges +3μC+3μC and +12μC+12μC are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force on a −2-nC charge when placed at the following locations: (a) halfway between the two (b) half a meter to the left of the +3μC+3μC charge (c) half a meter above the +12μC+12μC charge in a direction perpendicular to the line joining the two fixed charges

a)  F_total = -648 10⁻⁴ N

,  b)   F_total = - 312 N,  c)   F_total = (38.6 i ^ + 883.4 j ^) 10⁻⁴ N

Explanation:

To solve this exercise we will use Coulomb's law

         F = k q₁ q₂ / r²

and that forces are vectors

They indicate that the charge q₁ = 3 10⁻⁴ C and the charge q₂ = 12 10⁻⁴ C being separated d = 1 m

Let's place the reference system on the charge q₁

a) we place a third charge q₃ = 2 10⁻⁹ C at the midpoint

        r = d / 2

        r = ½ m

Since all charges have the same sign, the forces are repulsive, so the total force is

        F_total = F₁₃ - F₂₃

let's develop this expression

        F_total = k q₁ q₃ / r² - k q₂ q₃ / r²

        F_total =

let's calculate

         F_total = 9 10⁹  (3- 12) 10⁻⁴

         F_total = -648 10⁻⁴ N

b)  place the charge q₃ at a distance r = -0.5 m to the left of the charge q₁  

        F_total = -F₁₃ -F₂₃  

       F_total = -k q₁ q₃/ r₁₃² - k q₂ q₃ / r₂₃²  

      F_total = -  

the distances  

      r₁₃ = 0.5 m  

     r₂₃ = 0.5 +1 = 1.5 m  

calculate  

    F_total = - 9 10⁹ 2 10⁻⁹ (3 /0.5² + 12 /1.5²) 10⁻⁴  

    F_total = - 18 (12 + 5.33) 10⁻⁴  

    F_total = - 312 N

c) the charged q₃ is placed on the charge q₂ on the y axis.  

In this case the charge q₂ exerts a force on the y-axis on the charge q₃  

F₂₃ = k q₂ q₃ / r² j ^

where r = 0.5 m  

let's calculate  

F₂₃ = 9 10⁹ 12 10⁻⁴ 2 10⁻⁹ / 0.5²  

F₂₃ = 864 10⁻⁴ j ^  

Charge q1 exerts a repulsive force on the axis that joins them  

F₁₃ = k q₁ q₃ / r₁₃²

let's look for the distance  

r₁₃ =  

r₁₃ = 1,118 m  

Let's use trigonometry to find the direction of this distance  

tan θ = y / x  

θ = tan⁻¹ (y / x)  

θ = tan⁻¹ (0.5 / 1)  

θ = 26.6º

let's write this force in its components  

F₁₃ = k q₁  

(cos 26.6 i ^ + sin 26.6 j ^)  

let's calculate  

F₁₃ = 9 10⁹ 3 10⁻⁴  

(cos 26.6 i ^ + sin 26.6 j ^)  

F₁₃ = 43.2 (0.894 i ^ + 0.448 j ^) 10⁻⁴  

F₁₃ = (38.6 i ^ + 19.4 j⁾ 10⁻⁴ N  

Let's write the total force  

F_total = F₁₃ i ^ +  

j ^ + F₂₃ j ^  

let's substitute  

F_total = 38.6 10⁻⁴ i ^ + (19.4 + 864) 10⁻⁴ j ^  

F_total = (38.6 i ^ + 883.4 j ^) 10⁻⁴ N

d. learned preferences i believe is your answer

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.